Question

Both part of the question is True or False. Thank you

Problem 1. (ref. Example 3 in the slide) Let X = Y = C[0, 1] (with the norm || ||C[0,1] = sup |u(x)]). For any u € C[0, 1], define T€[0,1] v(t) = u(s)ds. We denote by T the mapping from u to v with D(T) = C[0, 1], i.e., v(t) = Tu(t). Then, the following conditions are true or not? Example 3. We denote by the set of all sequences u= (uk)k=1,2,. = (U1, U2, ...) (UK € R) (1) T is a functional. satisfying muxl < 00. We observe from luk + vkl? <244+ V13) (2) T is linear. that u, vel? => +ve. Hence, 12 is a vector space. Moreover, since (3) T is isometric. UKWK <2-(4x2 + 4x12), we see that for any u, V EB, Euk Vk converges absolutely. Thus, defining (1) as (4) T is a unitary operator. (u, v) = UK VK (5) T is bounded. we can verify that (: -) is an inner product of R. Then, B2 is a Hilbert space. (The method of the proof of completeness is (6) T is continuous. the same as that of l in the 5th lecture.) k1

$Problem 2. (ref. Example 4 in the slide) Let X = Y = C[0, 1] (with the norm ||$||C[0,1] = sup \u(x)]). For any u € C[0, 1],$

Answer 1

Tin not functional since it is Problem D G taking values in Cco, 1 as a functions in c[0,1]. Tin linear. Yes, it is trul let LER and u, vt col]. Then t T( Luto) s Quro).ds t t ar s Urs) ds (es) ds [Integral in linear J. 0 a Tlust TO). t 11 T(U) 11 uls).ds ds 11 t 2 دے رہ (sup xele, | u( 2 ) 2 x 1) dx t lullctolo' f llull t Also not onto, not T in a bijection, Hence tis not isometry. since T can be T is not a surfective map so unitary also. and 111111 (VI Yes id borended T ) رام continuous Yes, it Tin bounded ЭТ ТА

Grinen Y = [0, 1] x = & (6) llullcto, 1] - sup. x [6] lucxlt sup lu call. NE[o,i Now , T (ult) ) = u'(t) т is not a functional, since derivativne in linear, isometric since T it not oned one Q Tin linear 3 T is not 4 all T(o)= 0 constant functions in ( [0, 1] maps on Tin unitary oberator. T is not bounded . not ha n-1 VnEN. Tang= Ол no 809 - Tis an unbounded since Tin not continuary operators,